AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, 
, 
, 
lie on a semicircle 
in this order with diameter 
, and is not parallel to 
. Points 
and 
lie on segments 
and 
respectively such that 
and 
. A circle 
passes through and is tangent to , and intersects again at 
. Prove that the lines 
, and 
are concurrent.
, determine the smallest integer 
, such that it is possible to place numbers 
around a circle so that the sum of every consecutive numbers takes one of at most values.
a grid of 9*9, we choose a little square in of this grid three times, we can choose three times the same.
and its prime fatorization 
, its false derived is defined by 
Prove that there exist infinitely many naturals such that 
.
that satisfy![\[f(xf(y) + f(x)) = f(x)f(y) + 2f(x) + f(y) - 1,\]](http://latex.artofproblemsolving.com/4/7/f/47f094ce57fe35d339ec3b93152ec3459a22ec48.png)
for every 
, and 
for every 
and 
, such that 
.
be a set of 
points in a plane that lie within a disk of radius 
billion. Define a 
as picking a point 
and reflecting it across 's centroid. Does there always exist a sequence of at most 
moves after which all points of are contained in a disk of radius ?
such that for every real number 
, 
is a perfect square if and only if 
is a perfect square. Prove that 
is the square of a polynomial with real coefficients.
be positive integers such that 
and 
Prove that if 
and 
have the same prime divisors, then 
is a power of 2.
be non-empty subsets of the positive integers, and let 
denote their union. Prove that![\[\sum_{i=1}^m \sum_{(a,b)\in H_i^2}\gcd(a,b)\ge\frac1m \sum_{(a,b)\in S^2}\gcd(a,b).\]](http://latex.artofproblemsolving.com/1/c/d/1cdf561d3af92f524628e0724e0c5ee24d8d6a09.png)
Proposed by Dávid Matolcsi, Berkeley
. Initially, the numbers 
are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers 
and 
are on the board, then 
can be erased as 
, or 
as 
, or 
as 
.) is it possible to end with only one number remaining on the board?
such that
where we have the counterclockwise-oriented linear paths:
for ![$t\in[0,1]$](http://latex.artofproblemsolving.com/e/7/0/e70754c1efb07bd54438989e7e001bafb98b429a.png)
;
for ![$t\in[0,2]$](http://latex.artofproblemsolving.com/a/c/f/acf32b254b3a5ebc23c07d5ca0c5a708d6f764b7.png)
;
for .
, then this leads to:
;
;
.
holds inequality 
![\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\frac{1 - e^{-2} \cos\left(2\left(u + \tan u\right)\right)}
{1 - 2e^{-2} \cos\left(2\left(u + \tan u\right)\right) + e^{-4}}
\, \mathrm{d}u
\]](http://latex.artofproblemsolving.com/9/1/1/91165e6b531233c56b799ed32ded982d57b6c87a.png)
,
where the curve 
in the first octant, traversed in the counterclockwise direction as viewed from above, using the definition of line integral and the Stokes's theorem.
